For a finite sequence of homogeneous polynomials we also write for where .

Also let .

Exercise A

Prove the following, for any graded ideals and collection of graded ideals of B.

.

If is a set of homogeneous generators of , then

.

.

.

In the other direction, we define:

Definition.

Let be any subset. Then denotes the (graded) ideal of B generated by:

.

In summary, we defined the following maps.

Zariski Topology of Projective Space

We wish to define the Zariski topology on ; for that let us take subsets of which can be identified with the affine space . Fix ; let

Note that for the same point can be represented by where the i-th coordinate is 1. This gives a bijection . E.g. for n = 2, we have:

Note that for any , the intersection maps to an open subset of via both and . Indeed if i < j then is the set of all satisfying while is the set of all satisfying . Hence, the following is well-defined.

Definition.

The Zariski topology on is defined by specifying every as an open subset, where obtains the Zariski topology of from .

A projective variety is a closed subspace .

First, we have the following preliminary results.

Lemma 1.

For any homogeneous , the set is (Zariski) closed in .

Proof

It suffices to show that is closed in for each . But , where . The same holds for . ♦

Definition.

For any , let

be its homogenization.

Exercise B

1. Prove that if , the homogenization of fg is the product of the homogenizations of f and g.

2. Let be the homogenization of a non-constant . Then f is irreducible if and only if F is irreducible. [ Hint: you may find lemma 2 here helpful. ]

The Zariski topology on is consistent with our earlier notions of closed subsets:

Proposition 1.

A subset is closed under the Zariski topology if and only if for some graded ideal .

Proof

(⇐) Suppose for some homogeneous . Since , by lemma 1 this is Zariski closed.

(⇒) Let ; it suffices to show that any is contained in for some homogeneous . Now is contained in some , say without loss of generality. Hence for some . If , then

, where F = homogenization of f. ♦

Example

Let be the projective variety defined by the homogeneous equation . Then

is cut out from by ;

is cut out from by ;

is cut out from by .

Cone of Projective Variety

We wish to prove the bijective correspondence between graded radical ideals of B and closed subsets . For that, we can piggyback on existing results for the affine case.

Definition.

Let be any subset. The cone of C is

.

Note

For any non-empty collection of subsets we have

Also, we have:

Lemma 2.

If is a proper graded ideal then .

In particular, by proposition 1, the cone of a closed is closed in .

Proof

Note: if is non-constant homogeneous, then . Now pick a set of homogeneous generators for ; each is non-constant so

.

This completes the proof. ♦

Furthermore we have:

Lemma 3.

A non-empty closed subset is of the form for some closed if and only if

.

When that happens, we call W a closed cone in .

Proof

(⇒) is obvious; for (⇐) clearly W = cone(V) for some subset . Let so that . It remains to show is graded, for we would get by lemma 2.

Indeed if , write as a sum of homogeneous components. Then for any and we have which gives

Thus vanish for any , i.e. . ♦

Projective Nullstellensatz

Thus we have the following correspondences:

The top-left column is a bijection by lemma 3; the bottom row is a bijection by Nullstellensatz. In the proof of lemma 3, we also showed that for a closed cone , the ideal is graded. Conversely, if is graded, is the (non-empty) solution set of a collection of graded polynomials; hence it is a closed cone too.

Hence we have a bijection between

closed subsets , and

proper homogeneous radical ideals .

The correspondence takes and so

The last piece of the puzzle is the map which takes closed subsets of to homogeneous ideals of B. As an easy exercise, show that

.

However so we modify our bijection to:

Theorem (Projective Nullstellensatz).

There is a bijection between:

closed subsets ;

homogeneous radical ideals such that where is the irrelevant ideal of B.

Exercise C

Prove that if is a homogeneous ideal then is empty if and only if contains a power of .